Find the coefficient of x^3 in the expansion of (2x^2-3/x)^3 I know how to do simple coefficients using pascalles triangle but I really don't know how to do this. Any help would be much appreciated.
Write it as [tex](2x^2 - 3x^{-1})^3[/tex] From Pascal's triangle, you know how to expand [tex](a+b)^n[/tex] What can you replace with a and what can you replace with b?
Usually (well for me), a binomial expansion is usually done with a variable and a constant. as for [itex](a+b)^n[/itex] is valid for [itex]|\frac{b}{a}|<1[/itex] But if a and b are variables, you'll have to do some fancy algebra to get the range for which it is valid.
Why is it only valid in some range??? I also don't see why you need to factor the original. (a+b)^3=a^3+3*a^2*b+3*a*b^2+b^3. Just put a=2x^2 and b=(-3/x), figure out which term is the x^3 term and evaluate it.
Got a reference? If you are thinking of the convergence of the infinite series for negative exponents, that is something to think about. But this is a positive exponent, the series is finite. There are no convergence issues.
Yeah, that's what I mean by finite series. You could also just forget about pascal's triangle and multiply it out. The power is only 3.